JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

A circle touches

y

-axis at

\left( {0,2} \right)

and has an intercept of 4 units on the positive side of

x

-axis. The equation of the circle is

View answer and explanation

Correct option: A

{x^2} + {y^2} - 4\left( {\sqrt 2 x + y} \right) + 4 = 0

Option A (

{x^2} + {y^2} - 4\left( {\sqrt 2 x + y} \right) + 4 = 0

) is correct because it satisfies the given conditions of touching the y-axis and having a specific x-intercept.

More Questions from CIRCLES

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{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

x

\left( {\alpha ,2{\alpha ^2}} \right)

{x^2} + {y^2} = 4,

{x^2} + {y^2} - 2x + 4y - 4 = 0

{x^2} + {y^2} = {a^2}

\left( {{x_1},{y_1}} \right)

\quad \alpha, \beta

\gamma

P, Q

x^2+y^2=1

A

(-1,0)

A P, A Q

A R

G P

\cos \alpha / 2, \cos \beta / 2

\cos \gamma / 2

x+y=6

x+2 y=4

x+y=5

x-y=1

9 \pi